Let $DP$ meet $BQ$ at $P'$. Let $Q'$ be the rotation of $Q$ about the origin $90^\circ$ in the direction from $A$ to $B$. Observe $P'$ lies on $(ABCD)$, whence one can easily show $AQ = QP'$ and $P'Q' = Q'C$. Also note $Q', A, R$ collinear. From similarity we are done since $AQ, DP, CQ'$ are all parallel.

I'm new to aops, so I couldn't figure it out how to attach an image. Sorry about that Let's construct point $M$ outside the square such that triangles $BMC$ and $AQB$ are equal in that order. By angle chasing, $M$ lies on $BQ$ and on $BR$. But now $AR$, $CQ$ and $DP$ are three altitudes for truiangle $DQM$. Hence, we are done.

Since $\angle ABQ=\angle ADP$ and $\angle AQB=90^{\circ}=\angle DAP$, $\angle APD=\angle QAB$, i.e. $AQ \parallel DP$. Furthermore, since $\angle BAR=\angle ADQ$, $\angle QDR=\angle ARC$, which implies that $D,C,R,AR \cap DQ$ are concyclic. So we also have $AR \perp DQ$. We will use Cartesian coordinates to complete the rest of our solution. Set $A(0,0),B(1,0),C(1,-1),D(0,-1)$ and let $P$'s coordinates be $(a,0)$. Since $AQ \parallel DP$ and $m_{DP}=\frac{1}{a}$, the equation of line $AQ$ is $y=\frac{1}{a}x$. Since $BQ \perp AQ$, $BQ$'s slope is $-a$, and therefore, its equation is $y=-ax+a$. Then, $Q=AQ \cap BQ$'s coordinates are $\left(\frac{a^2}{a^2+1},\frac{a}{a^2+1}\right)$. It follows that the slope of line $DQ$ is $\frac{a^2+a+1}{a^2}$. Since $DQ \perp AR$, line $AR$ 's slope is $-\frac{a^2}{a^2+a+1}$, and its equation is $y=-\frac{a^2}{a^2+a+1}x$. The equation of line $DP$ is $y=\frac{1}{a}x-1$, so the coordinates of $AR \cap DP$ are $\left(\frac{a^3+a^2+a}{a^3+a^2+a+1},\frac{-a^3}{a^3+a^2+a+1}\right)$. Its suffices to show that this point lies on line $CQ$. Clearly, it lies on $CQ$ as $\frac{\frac{-a^3}{a^3+a^2+a+1}-(-1)}{\frac{a^3+a^2+a}{a^3+a^2+a+1}-1}=-a^2-a-1=\frac{\frac{a}{a^2+1}-(-1)}{\frac{a^2}{a+1}-1}$. $_\blacksquare$

构造弦图.

Attachments:
2024AustralianMO-P7-2.pdf (88kb)

Let $DP$ meet $ABCD$ at $T$. Let $AR$ and $BQ$ meet at $S$.Let $DP$ meet $AR$ at $K$. Note that $\angle TBA = \angle PDA = \angle ABQ$ so $T$ lies on $BQ$. Also since $PT \perp BQ \perp AQ$ we have that $PT \parallel AQ$. Note that $\angle QDA = \angle BAS$ and $\angle SBA = 180 - \angle ABQ = \angle DAQ$ and $DA = AB$ so $DAQ$ and $BAS$ are congruent. Since $BS = AQ$ and $\angle SBC = \angle QAB$ then $SBC$ and $QAB$ are congruent. Since $AQ \parallel CS$, we need to prove $\frac{AK}{KS} = \frac{AQ}{SC}$. Note that $\frac{AK}{KS} = \frac{QT}{TS} = \frac{AQ}{SC}$ as wanted.